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I am using LM2596 for a dc supply where I could alter the output voltage. Datasheet here.

I have setup according to page 9 "Adjustable Output Voltage Versions". Everything worked fine. The output voltage alters as I changes the value of R2 (potentiometer used for R2, R1 remains 1K).

enter image description here

Now, I was thinking that the feedback pin is just taking a the voltage potential between R1 and R2. Thus instead of using R1 and R2, I supply a voltage (0-2V) from the bench power supply directly to the feedback pin. But as I vary the bench power supply, the output voltage does not vary. The power supply and the chip is common grounded.

Why is that? When using R1 and R2, the voltage on the feedback pin is around 1.3V. Removing R1 and R2, and supply feedback pin with a 1.3V does not yield the same output voltage. Or must there be a certain current flow?

The final destination would be to digitally control the chip (e.g. using a digital potentiometer or DAC).

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    \$\begingroup\$ "The final destination would be to digitally control the chip (e.g. using a digital potentiometer or DAC)." Then I urge you to study some circuit designs of digitally controlled regulator circuits, just to have an example of how this is done, it works the same with linear (non switching) regulators so also look at these. Usually these use an additional opamp to do the actual voltage regulation and use the voltage from a DAC as the reference voltage. It can be done but it requires some knowledge on analog electronics ! \$\endgroup\$ – Bimpelrekkie Oct 9 '15 at 8:51
  • \$\begingroup\$ An example of a schematic of a microcontroller controlled power supply using a simple switching regulator is here: eevblog.com/forum/reviews/… Note that the DAC here is very simple, a PWM output is used and filtered (a poor man's DAC so to say). When you understand this schematic, you're ready to design your own ;-) \$\endgroup\$ – Bimpelrekkie Oct 9 '15 at 11:17
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It's a feedback loop. It will vary the output and read the output voltage, compare that with the internal reference and set the output voltage one way or the other until the output voltage divided by the resistors matches the internal reference.

What you have done is cut the loop. The regulation of the output will have no effect on the input to the feedback pin, so it will basically do all sorts of stuff, if you hit the internal reference exactly it should stop changing the output, but that is only a theoretical case.

If you want to regulate your output you can inject a current on the bottom feedback resistor, but you need to have the resistor divider still connected:

schematic

simulate this circuit – Schematic created using CircuitLab

Depending on the choice of R1 and R3 you get a settable range of the output.

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  • \$\begingroup\$ Just want to ask, the value of R2 doesn't matter? How can I calculate the output per bench supply value? E.g. Given R2 = 1kohm. When the bench supply is at 0V, then R1 and R3 are in parallel, thus Vout = (1 + 1000/500) * 1.235V = 3.705V? But say when the bench supply is at 3.3V... how should the calculation work? [the internal ref of the LM2596 is 1.235V] \$\endgroup\$ – MW_hk Oct 19 '15 at 8:07
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The voltage at the feedback pin is typically 1.23 volts but can be as low as 1.18 volts or as high as 1.28 volts (see top of page 4 of the data sheet). Somewhere in that small range will be a voltage that, if exceeded by a millivolt or so, will turn the output off completely. If you were a milli volt under that voltage, the output would turn on completely and produce an output voltage limited only by the incoming power supply rail.

So meddling with it comes with a big caution because you can easily destroy the target circuit due to over voltage. Think of it like a high gain amplifier that within the range of 1 milli volt (or thereabouts) the output can swing between 0V and Vsupply.

You definitely need that feedback connection intact but you can force a current into the node with a little bit of care - maybe use a pot and an op-amp unity gain buffer feeding the node thru a 10k resistor to experiment with - this should provide +/-10% variation and if you get confident you can lower the 10k to maybe 2k2 and experiment to see what happens.

Be prepared for instability though if you push it too far.

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    \$\begingroup\$ A relatively safe method of manipulating the output voltage of a (switching) converter through the FB pin is by using a diode ! Place the diode such that it can only pull the FB pin UP in voltage. Then anything you do through the diode can only LOWER the output voltage ! This scheme is often used to add adjustable current limiting to a regulator chip that has no provisions for this. \$\endgroup\$ – Bimpelrekkie Oct 9 '15 at 11:04
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    \$\begingroup\$ An example, This module: ebay.nl/itm/… uses this scheme, schematic here: eevblog.com/forum/reviews/… \$\endgroup\$ – Bimpelrekkie Oct 9 '15 at 11:14

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